Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main()
{ int n;
cin>>n;
double p[n];
for(int i=0;i<n;i++) cin>>p[i];
double dp[n][n+1];
dp[0][0]=1-p[0];
dp[0][1]=p[0];
for(int i=1;i<n;i++)
for(int j=0;j<=i+1;j++) 
if(j==0) dp[i][0]=(1-p[i])*dp[i-1][0];
else dp[i][j]=p[i]*dp[i-1][j-1]+(1-p[i])*dp[i-1][j];

double ans=0;
for(int i=ceil(n/2.0);i<=n;i++) ans+=dp[n-1][i];

cout<<setprecision(10)<<ans;
return 0;
}